Introduction to Complex Analysisby Hilary Priestley

Unofficial Solutions Manual

MOHAMMAD EHTISHAM AKHTAR

IMPERIAL COLLEGE LONDON

http://akhtarmath.wordpress.com

Dedicated to my Parents

ii

Preface

This is an ongoing Solutions Manual for Introduction to Complex Analysis by Hilary Priestley[1]. The main reason for taking up such a project is to have an electronic backup of my ownhandwritten solutions.

Mathematics cannot be done without actually doing it. However at the undergraduatelevel many students are put off attempting problems unless they have access to written so-lutions. Thus I am making my work publicly available in the hope that it will encourageundergraduates (or even dedicated high school students) to attempt the exercises and gainconfidence in their own problem-solving ability.

I am aware that questions from textbooks are often set as assessed homework for stu-dents. Thus in making available these solutions there arises the danger of plagiarism. Inorder to address this issue I have attempted to write the solutions in a manner which con-veys the general idea, but leaves it to the reader to fill in the details.

At the time of writing this work is far from complete. While I will do my best to addadditional solutions whenever possible, I can not guarantee that any one solution will beavailable at a given time. Updates will be made whenever I am free to do so.

I should point out that my solutions are not the only ways to tackle the questions. It ispossible that many ‘better’ solutions exist for any given problem. Additionally my workhas not been peer reviewed, so it is not guaranteed to be free of errors. Anyone using thesesolutions does so at their own risk.

I also wish to emphasize that this is an unofficial work, in that it has nothing to do withthe original author or publisher. However, in respect of their copyright, I have chosen toomit statements of all the questions. Indeed it should be quite impossible for one to read thiswork without having a copy of the book [1] present.

I hope that the reader will find this work useful and wish him the best of luck in hisMathematical studies.

MOHAMMAD EHTISHAM AKHTAR

IMPERIAL COLLEGE LONDON

Project started on 29 June 2008

The end of a solution is indicated by �. Any reference such as ‘Theorem 13.1’, ‘Question23.10’ refers to the relevant numbered item in Priestly’s book [1]. This work has been pre-pared using LATEX.

The latest version of this file can be found at : http://akhtarmath.wordpress.com/

Cite this file as follows :

Akhtar, M.E. Unofficial Solutions Manual for Introduction to Complex Analysis by H.A.Priestley.Online book available at : http://akhtarmath.wordpress.com/

Contents

Preface ii

Quick Reference 2

1 The Complex Plane 3

2 Geometry in the Complex Plane 5

3 Topology and analysis in the complex plane 7

5 Holomorphic functions 9

6 Complex series and power series 11

Bibliography 13

Conditions of Use

This work is Copyright c© 2008 Mohammad Ehtisham Akhtar. You are permitted to use anddistribute this work freely for non-profit purposes as long as you do not modify it in anyway. If you use this work or quote parts of it then you must give a proper citation. The useand distribution of this work for commercial purposes is prohibited.

2 Contents

Quick Reference

Chapter 1: The Complex Plane

1.12 , 1.13 , 1.14 ,

Chapter 2: Geometry in the Complex Plane

2.4 , 2.14 ,

Chapter 3: Topology and analysis in the complex plane

3.9 ,

Chapter 5: Holomorphic functions

5.1 , 5.3 , 5.4 , 5.6 , 5.7 ,

Chapter 6: Complex series and power series

6.3 , 6.4 , 6.7 ,

Chapter 1

The Complex Plane

1.12) Given z, w ∈ C we have that :

|1− zw|2 − |z − w|2 = (1− zw)(1− zw)− (z − w)(z − w)

= 1 + |z|2|w|2 − |z|2 − |w|2 =(1− |z|2

) (1− |w|2

)as required. Now suppose |z| < 1 and |w| < 1. We observe that :∣∣∣∣ z − w1− zw

∣∣∣∣ < 1 ⇐⇒ |z − w| < |1− zw| ⇐⇒ |z − w|2 < |1− zw|2

where the last equivalence follows because both |z − w| ≥ 0 and |1 − zw| ≥ 0. Furthermore|z − w|2 < |1 − zw|2 if and only if

(1− |z|2

) (1− |w|2

)> 0 by the previous part. This last

inequality is true because both |z| < 1 and |w| < 1 by assumption. The result follows. �

1.13) (i) Recall that for a complex number a, we have 2Re(a) = a + a, and if a 6= 0 then(a−1) = (a)−1. With these ideas in mind, select z, w ∈ C with z 6= w. Then :

Re(w + z

w − z

)=

12

[(w + z

w − z

)+(w + z

w − z

)]=

12

[(w + z

w − z

)+

(w + z)(w − z)

]

=12

[(w + z

w − z

)+

(w + z)(w − z)(w − z)(w − z)

]

=12

[(w + z)(w − z) + (w + z)(w − z)

|w − z|2

]=

12

[2(|w|2 − |z|2

)|w − z|2

]=|w|2 − |z|2

|w − z|2

as required. (ii) We observe that |w − z|2 = (w − z)(w − z) = |w|2 − 2Re(zw) + |z|2. Writingz = reiθ and w = Reiϕ we find that |w − z|2 = R2 − 2Re

(Rrei(θ−ϕ)

)+ r2 and since eiα =

cos (α) + i sin (α) for α ∈ R we conclude that |w − z|2 = R2 − 2Rr cos (θ − ϕ) + r2. Bysubstituting this result, along with |z| = r and |w| = R into the expression obtained Part (i)we deduce that

Re(w + z

w − z

)=

R2 − r2

R2 − 2Rr cos (θ − ϕ) + r2

which was what we wanted to show. �

1.14) Suppose there exists a relation > on C satisfying both (a) and (b). Since i 6= 0, condition(a) implies that exactly one of i > 0 or −i > 0 holds. If i > 0 then condition (b) implies that

4 1. The Complex Plane

i2 = −1 > 0. Now1 since both i > 0 and −1 > 0, it follows once again from condition (b)that(−1)(i) = −i > 0. But now both i > 0 and −i > 0, which contradicts condition (a). On theother hand, if −i > 0 then we once again find that (−i)2 = −1 > 0 which, using condition(b) implies (−1)(−i) = i > 0. This once again contradicts condition (a) as before. �

1Note that the fact that−1 > 0 does not in itself yield a contradiction. This is because > may not be the usualorder relation on R.

Chapter 2

Geometry in the Complex Plane

2.4) We shall use the same notation as in [1] : (i) Here α = −i, β = 3i and λ = 1. The givenequation describes the straight line Im(z) = 1 in C. (ii) Here α = −1, β = 1 and λ = 4.Since λ > 0 and λ 6= 1, the given equation describes a circle in C. The diameter of this circlethrough α and β intersects the circle at z1 and z2 where z1 − α = λ(z1 − β) ⇐⇒ z1 + 1 =4(z1 − 1) and z2 − α = −λ(z2 − β) ⇐⇒ z2 + 1 = −4(z2 − 1). We conclude that z1 = (5/3)and z2 = (3/5). Therefore, the centre of the circle is : (1/2)(z1 + z2) = (17/15) and the radiusof the circle is : (1/2)|z1 − z2| = (8/15). (iii) Here α = i, β = 0 and λ = 2. Proceeding as in(ii), the equation describes a circle in C centered at (−i/3) with radius (2/3). (iv) Here α = 0,β = i and λ = 2. Proceeding as in (ii), the equation describes a circle in C centered at (4i/3)with radius (2/3). The sketches are left to the reader. �

2.14) Remark. There seem to be two misprints in this question as stated in [1]. The first is inpart (ii)(b) : The equation of the arc is probably meant to be arg((z−α)/(z−β)) = µ (mod 2π).The second is in part (iii) : K should be c/(a− cα), not 1/(a− cα).(i) A complex number z ∈ C is a fixed point of the Mobius transformation f if and only ifz = f(z) = (az + b)/(cz + d), if and only if z is a root of the polynomial : cz2 + (d − a)z − b(a, b, c, d ∈ C) The stated polynomial has either one or two roots in C ; Therefore, f has eitherone or two fixed points.(ii) Suppose that α, β ∈ C are two distinct fixed points of f . The inverse map of f is g :z 7→ (dz − b)/(a − cz). Since f(α) = α and f(β) = β, it must be the case that α = g(α) =(dα− b)/(a− cα) and β = g(β) = (dβ − b)/(a− cβ). Using these facts we observe :

w − αw − β

=(a− cα)z − (dα− b)(a− cβ)z − (dβ − b)

=(a− cα)a− cβ

[z − g(α)z − g(β)

]= k

z − αz − β

where k = (a − cα)/(a − cβ) as required. From this we see that w−αw−β = 1

k

(z−αz−β

). Therefore,

(a) The image of the given circline under f is the circline : |(w − α)/(w − β)| = |k|λ. For (b)we assume that the arc is described as in the remark above. Then the image of the arc underf is the arc : arg((w − α)/(w − β))− arg(k) = µ (mod 2π).(iii) Suppose that f has a single fixed point α. Then α is a repeated root of the polynomialstated in part (i). That is : α = (a − d)/2c. Equivalently, d = a − 2cα. Also, as in part(ii),

6 2. Geometry in the Complex Plane

g(α) = α. Using these facts we observe :

1w − α

=cz + d

(a− cα)z − (dα− b)

=cz + a− 2cα

(a− cα)(z − g(α))=

1z − α

+K

where K = c/(a− cα) as required (see the remark at the beginning of this solution). �

Chapter 3

Topology and analysis in the complex plane

3.9) (a)(i) Since |zn| = (1/n) → 0, the sequence {(1/n)in} converges to 0. (a)(ii) Note that|1 + i| =

∣∣∣√2eiπ4

∣∣∣ =√

2. Therefore |zn| =(1/√

2)n → 0 and so the sequence {(1 + i)−n}

converges to 0. (a)(iii) Since (n2 + in)/(n2 + i) =(1 + in−1

)/(1 + in−2

), the sequence{

(n2 + in)/(n2 + i)}

converges to 1. (b)(i) {in} does not converge because it contains thedivergent sequence {(−1)n} as a subsequence. (b)(ii) Since |zn| = |(1 + i)n| =

(√2)n → ∞,

it follows that {(1 + i)n} does not converge. (b)(iii) The real part of {n(−1)n/(n + i)} is thedivergent sequence

{n2(−1)n/

(n2 + 1

)}. Thus the original sequence does not converge. �

Chapter 5

Holomorphic functions

5.1) (i) For any z = x + iy ∈ C, Im z = u(x, y) + iv(x, y), where u(x, y) = 0 and v(x, y) = y

for all (x, y) ∈ R2. In particular ux = 0 6= 1 = vy at any (x, y) ∈ R2. So the Cauchy-Riemannequations fail at every z = x+ iy ∈ C. We conclude Im z is not differentiable anywhere.(ii) For any z = x+ iy ∈ C, z = u(x, y) + iv(x, y), where u(x, y) = x and v(x, y) = −y for all(x, y) ∈ R2. In particular, ux = 1 6= −1 = vy. So the Cauchy-Riemann equations fail at everyz = x+ iy ∈ C. We conclude z is not differentiable anywhere. �

5.3) If f = u+iv is differentiable at z = a+ib ∈ C, then all the partial derivatives ux, uy, vx, vyexist at z, and in particular, vx = −uy there. Now f ′(z) = ux + ivx and also f ′(z) = vy − iuy.Substituting vx = −uy into both of these yields f ′(z) = ux − iuy and f ′(z) = vy + ivx. �

5.4) In each part of this question, we shall denote the given function by f for convenience.(i) If h 6= 0 then 1

h(f(0 + h) − f(0)) = |h|2/h = h → 0 as h → 0. Therefore, f(z) = |z|2 isdifferentiable at z = 0 and f ′(0) = 0. (ii) If z = x+ iy then f(z) = x+y. Therefore, f = u+ iv

where u(x, y) = x + y and v(x, y) = 0. If f is differentiable at z = 0, then all the first partialderivatives of u and v must exist at z = 0 and must satisfy the Cauchy-Riemann equationsat z = 0. However, ∂u

∂x(0, 0) = 1 6= 0 = ∂v∂y (0, 0), and thus we conclude that the function

f is not differentiable at z = 0. (iii) If z = x + iy then f(z) = xy. Therefore, f = u + iv

where u(x, y) = xy and v(x, y) = 0. One can check that all the first partial derivatives of uand v are continuous in C and that they satisfy the Cauchy-Riemann equations at z = 0 ∈ C.Therefore, the function f(z) = (Re z)(Im z) is differentiable at z = 0. �

5.6) Suppose that there exists c ∈ [1, i] such that the given equation holds. This is equivalentto saying that there exists t ∈ [0, 1] such that c = (1− t) + it and (1 + i)/(1− i) = 3c2. Takingmodulus on both sides, the second equation implies that :∣∣∣∣1 + i

1− i

∣∣∣∣ = 3|c|2 ⇒ |c| = 1√3

Now also c = (1−t)+it for some t ∈ [0, 1]. So√

1− 2t+ 2t2 = |c| = 1/√

3, which implies thatt must be a root of the quadratic : 6t2 − 6t + 2. But since the discriminant (−6)2 − 4(6)(2) isnegative, tmust have nonzero imaginary part. This contradicts the assumption that t ∈ [0, 1].Therefore, there exists no c ∈ [1, i] satisfying the given equation. �

5.7) (a)(i) This is a polynomial and so is holomorphic at all z ∈ C. (a)(ii) Holomorphic at allz ∈ C except at z = 0, 1 and 2. (a)(iii) Holomorphic at all z ∈ C except at z = 1, ω, ω2, ω3

and ω4, where ω = exp (2πi/5). (b)(i) First select any a ∈ C\{0}. If 1/|z| is holomorphicat a then in particular 1/|z| is differentiable at a. Also 1/|a| 6= 0 so (1/|z|)−1 = |z| is also

10 5. Holomorphic functions

differentiable at a. This is a contradiction since |z| does not satisfy the Cauchy-Riemannequations at a (6= 0). So 1/|z| is not holomorphic at any a ∈ C\{0}. It follows that 1/|z| cannot be holomorphic in any disc centered at 0. Thus 1/|z| is not holomorphic at any a ∈ C.(b)(ii) Suppose that z|z| is holomorphic at a ∈ C\{0}. The function 1/z is also holomorphicat a (6= 0). Therefore (1/z)(z|z|) = |z| is also holomorphic at a. This is a contradiction since|z| is not even differentiable at a. The argument now follows as in (b)(i). �

Chapter 6

Complex series and power series

6.3) Viewing∑zn as a power series expansion of (1− z)−1 for |z| < 1, we have :

(i)(a) 11−z = 1

2(1− (z+1)

2

) =∑∞

n=0(z+1)n

2n+1 ; this converges for |z + 1| < 2.

(i)(b) 11−z = 1

(1−i)(1−( z−i1−i))=∑∞

n=0(z−i)n

1−in+1

; this converges for |z − i| < |1− i| =√

2.

(ii)(a) 1z(z+2) = 1

z2+2z= 1

1+(z+1)2=∑∞

n=0(−1)n(z + 1)2n ; this converges for |z + 1| < 1.

(ii)(b) For this part we first express 1/z and 1/(z+2) individually as power series in powersof z − i, and then use the multiplication rule for power series. The resulting series willconverge for min {r1, r2}, where r1 and r2 are the radii of convergence of each of the serieswe have just obtained. Hence :

1z(z + 2)

=

(−i

∞∑n=0

in(z − i)n)·

(1

2 + i

∞∑m=0

(−1)m(z − i)m

(2 + i)m

)=−i

2 + i

∞∑n=0

cn(z − i)n

where cn =∑n

k=0in−k(−1)k

(2+i)k. This series will converge for |z − i| < min {1,

√5} = 1 . �

6.4) (i) By the ratio test, the series has infinite radius of convergence, and thus1 it definesa holomorphic function at every point of C. (ii) By the ratio test, the series has radius ofconvergence equal to 1, and thus it defines a holomorphic function on the open unit disc.(iii) By Cauchy’s root test, the series has infinite radius of convergence, and thus it defines aholomorphic function on C. (iv) By the ratio test, the series has radius of convergence equalto 0. Therefore, it does not define a holomorphic function at any point2 of C. �

6.7) By the differentiation theorem for power series :(i) (1 + z)−2 = − d

dz (1 + z)−1 = −∑∞

n=1 (−1)nnzn−1 =∑∞

n=1 (−1)n+1nzn−1 ; and(ii) (1 + z)−3 = −(1/2) ddz (1 + z)−2 = (1/2)

∑∞n=2 (−1)nn(n− 1)zn−2

where both of the obtained expansions are valid for |z| < 1. �

1Using the differentiation theorem for power series2Since such a function will not even be defined, let alone differentiable, in an open disc about any point of C.

Bibliography

[1] Priestley, H.A. Introduction to Complex Analysis, 2nd Ed., 2003. Oxford University Press.